Blinker circuit issue

Discussion in 'General Electronics Chat' started by CTRL+X, Mar 8, 2015.

  1. CTRL+X

    Thread Starter Member

    Jul 1, 2009
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    Seeing if someone can diagnose the problem with this. Trying to make a simple led blinkers circuit.

    Attempting to duplicate the circuit here... http://cappels.org/dproj/simplest_LED_flasher/Simplest_LED_Flasher_Circuit.html

    The led always remains on. Increasing the resistor or decreasing the input voltage only results in the led dimming and with sufficient resistance not turning on at all.
    It should charge the capacitor until the voltage is sufficient to cross the transistor, discharge, recharge the capacitor ect.....?
    Can someone point out what I am doing wrong?

    Thanks,

    [​IMG]
     
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    What are you using to power the circuit?
    Is the transistor orientated as described?

    Bertus
     
  3. CTRL+X

    Thread Starter Member

    Jul 1, 2009
    26
    0
    Powering with 12v dc. I believe the transistor is oriented correctly, however I have tried switching it around and the led doesn't light up at all.
     
  4. bertus

    Administrator

    Apr 5, 2008
    15,648
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    Hello,

    What color led do you use?
    In the original circuit a red one is used.
    A white one will not work, unless you try a 15 Volts powersupply.

    Bertus
     
  5. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,810
    834
    What are the resistor values? The one used to charge the resistor doesn't look like a 1k (could be the picture) I don't see a red band. The original article states
    "If the resistor that charges the capacitor is too low in value (or if the power supply voltage is too high), the current through the transistor will not become low enough for the transistor to turn off. "
    Which is the symptom you describe.
     
  6. CTRL+X

    Thread Starter Member

    Jul 1, 2009
    26
    0
    I am using a red led.
    Resistor values are 1k connecting to + and 100 ohm from the led to ground.
    Yeah, the blue resistor color makes it almost impossible to see the band colors, but I tested both with a multimeter to verify.

    I also tried duplicating the 2 transistor circuit shown a little further down the page, and ran into the same issue.
     
  7. CTRL+X

    Thread Starter Member

    Jul 1, 2009
    26
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    According to the specs for 2N2222A transistor which I am using, the breakdown voltage across collector and emitter with the base floating is 40V. So I don't quite understand why the current is bridging in this case, or even why the concept works as the schematic describes, with the cap building sufficient charge to allow the current to flow.
    I don't need the circuit for anything. Just trying to understand the concept, and why it isn't working.
     
  8. GopherT

    AAC Fanatic!

    Nov 23, 2012
    6,061
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    @DickCappels , the owner of the website you referenced, will be here shortly to help. Also, post another photo showing how your battery is connected.
     
  9. CTRL+X

    Thread Starter Member

    Jul 1, 2009
    26
    0
    Attached is a photo showing the battery jumpers to the board. Also a picture of the battery pack. It is running off of 8 1.5v cells.
    As shown below, the led is lit.
    I've looked as this circuit all day, and swapped out multiple components to no avail. It's got to be something simple I'm missing.


    [​IMG]

    [​IMG]
     
  10. DickCappels

    Moderator

    Aug 21, 2008
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    From first appearance, it looks like either your power supply is too high or your 1k resistor is too low or both. The characteristics upon which the oscillator depends is not specified by the manufacturer. Some transistors are "finicky" and need to have the voltage and timing resistor carefully adjusted in order for oscillation to occur. By the way, the conditions for oscillation can vary significantly from batch-to-batch.

    In this circuit, the transistor changes between a "latched-on" state and an "off" state. Assuming that the timing capacitor is discharged when we first look at the circuit, there will be about 12 volts (assuming a 12 volt power supply) across the timing resistor and zero volts across the transistor. The voltage at the emitter will slowly increase until the emitter-base junction becomes sufficiently revers biased and supplied with enough current for it to avalanche. To achieve this the voltage must be significantly higher than the reverse breakdown voltage of the emitter-base junction plus the forward biased collector-base junction. If there is sufficient current through the timing resistor to provide the current necessary to get the transistor into avalanche mode.

    When the voltage on the capacitor gets to its lowest voltage, the transistor needs to become starved of current sufficient that the transistor can turn off so the current thought the timing resistor must be small enough to allow this to happen.

    The two statements above tells us that there is a limited range of combinations of timing resistance and supply voltages that will work.

    Note for CTRL+X: The critical voltage occurs just above the emitter-base breakdown voltage + the voltage drop across the forward biased collector-base junction. The collector-to-emitter breakdown voltage is not applicable because in this circuit the emitter is acting as a collector and the collector is acting as an emitter.
     
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  11. CTRL+X

    Thread Starter Member

    Jul 1, 2009
    26
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    Thanks for the very detailed explanation. I will take a closer look at those perimeters and see if I can get it dialed in.
     
  12. ScottWang

    Moderator

    Aug 23, 2012
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    I just want to make sure that the circuit was really worked ... :)

    I did some test, the values may not precisely, because the values were changing all the time and I used 2sc1384, the values as below were what I recorded.

    Transistor : 2SC1384, C : 220uf.
    LED:2V/20mA/5mm, 1K,100Ω are the same.

    Simplest LED Flasher Circuit_1bjt.gif
     
  13. Dodgydave

    Distinguished Member

    Jun 22, 2012
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  14. ScottWang

    Moderator

    Aug 23, 2012
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  15. Roderick Young

    Member

    Feb 22, 2015
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    I don't think that the "equivalent circuit" above would work as a flasher. In the case of a transistor, extra holes can get into the base region, resulting in positive feedback, and negative resistance. The circuit with the Zener wouldn't inject holes into the junction of the diode below and make it conduct. It's just like, one can't just connect two diodes and make a transistor. The circuit may read like a transistor on an ohmmeter, but will not amplify.
     
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  16. CTRL+X

    Thread Starter Member

    Jul 1, 2009
    26
    0
    I messed around with this a bit yesterday, but all was able to get was either the LED to dim, or induce a delay in lighting up.

    Lowering the voltage, raising the 1k resistor value, or combo of the two - results in the LED dimming, or with low V or high enough R, not lighting up.

    Raising the capacitor size at C1 - Causes a delay correspondent to the cap size in the LED lighting up, but then stays lit up. Discharging the cap induces the delay again, but once charged the delay does not occur.
    This leads me to believe that the initially the circuit is working. The voltage remains low enough when charging the cap, that it doesn't pass the transistor.
    However once charged, the cap just remains as is, and doesn't discharge.
     
  17. ScottWang

    Moderator

    Aug 23, 2012
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    You are right.
    The equivalent circuit just to make it more easy to understand how the circuit works, but not really can using Zener to replace it, we also used that two diodes to explain the same thing.
     
    Last edited: Mar 10, 2015
    Roderick Young likes this.
  18. Roderick Young

    Member

    Feb 22, 2015
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    Another crazy idea - what happens if you leave out the 100 ohms, that is, put a short in its place? Decades ago, we made flashers out of neon bulbs, using a very similar circuit (except that the supply had to be 100 volts or more). If there was too much resistance in series with the bulb, the lamp would just lock "on" dimly rather than oscillate.
     
  19. ScottWang

    Moderator

    Aug 23, 2012
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    Did you design the circuit and for what reason?
    Do you think will it more easy to damaged than the normal method as c connecting to positive and e connecting to negative?
     
  20. ScottWang

    Moderator

    Aug 23, 2012
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    There is an error shown in the #12 that is the frequency, it should be Hz and not the KHz, that's why I felt very strange if the frequency is in KHz area then how can I saw it flashing, so because that where I stand far from the O'scope caused that.

    Simplest LED Flasher Circuit_1bjt-02.gif
     
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